Prove that a polynomial ring is integrally closed 
Let $V \subseteq {\mathbb{A}}^2_{\mathbb{C}}$ be the curve defined by $x^2-y^2+x^3=0$, and let $\mathbb{C}\left [ V \right ]$ the coordinate ring of $V$. Let $\Theta :=\bar{y}/\bar{x} \in \mathbb{C}\left ( V \right )$. I must show that the ring $B:=\mathbb{C}\left [ V \right ]\left [ \Theta  \right ]$ is a UFD.

I could show that this ring is a FD, because it is noetherian, but I am not sure how to prove that the factorizations are unique. Any help would be appreciated. I also tried to show that $B$ is isomorphic to a UFD, but i am not sure what domain would be suitable for this argument.
(Note: I need to prove that $B$ is a UFD to say that $B$ is integrally closed.)
 A: Let us show that the $\Bbb{C}[V][\Theta]$ is in fact generated by $\Theta$ as a $\Bbb{C}$-subalgebra of $\Bbb{C}(V).$
Define a morphism $\phi$ as follows:
\begin{align*}
\phi : \Bbb{C}[x,y]&\to\Bbb{C}[t]\\
x&\mapsto t^2 - 1,\\
y&\mapsto t^3 - t.
\end{align*}
It is not difficult to check that this factors through the quotient map $\Bbb{C}[x,y]\to\Bbb{C}[V].$
Now, clearly $\phi$ is a surjection onto $\Bbb{C}[t^2 - 1,t^3 - t],$ so that $\Bbb{C}[x,y]/\ker\phi\cong\Bbb{C}[t^2 - 1,t^3 - t].$ We need to prove that $\ker\phi = (x^3 + x^2 - y^2)$. To do so, note that both $\ker\phi$ and $(x^3 + x^2 - y^2)$ are prime. Since $\ker\phi$ is not maximal ($\Bbb{C}[x,y]/\ker\phi$ is visibly not a field), it is properly contained in some maximal ideal $\mathfrak{m}.$ This gives us a chain of prime ideals
$$
(0)\subsetneq (x^3 + x^2 - y^2)\subseteq \ker\phi\subsetneq\mathfrak{m}.
$$
But, $\dim\Bbb{C}[x,y] = 2,$ so that we must have $(x^3 + x^2 - y^2) =  \ker\phi.$
Thus, we get an isomorphism
$$
\Bbb{C}[x,y]/(x^3 + x^2 - y^2)\cong\Bbb{C}[t^2 - 1,t^3 - t],
$$
and $y/x = \Theta$ in the fraction field of the left hand side corresponds to $t$ in the fraction field of the right hand side, because $x = \Theta^2 - 1$ and $y = \Theta^3 - \Theta.$ Now, it is easy to see the result, as $\Bbb{C}[V][\Theta]\cong\Bbb{C}[t^2 - 1,t^3 - t][t] = \Bbb{C}[t].$ This implies that $\Bbb{C}[V][\Theta] = \Bbb{C}[\Theta]$ (and that $\Theta$ satisfies no relations over $\Bbb{C}$) as claimed initially.

Edit: As user26857 notes, the initial solution I presented (below) is not totally rigorous -- we need some condition on $x$ and $y$ to guarantee that $R[\frac{y}{x}]\cong R[T]/(xT - y).$ In fact, it isn't true that $\Bbb{C}[V][T]/(xT-y)\cong\Bbb{C}[V][y/x]$: the ideal $(xT-y)$ should be $(xT -y, T^2 - x - 1)$ -- this second relation is implicitly assumed and explicitly used. What is below can be made rigorous, either by justifying that the kernel of $\Bbb{C}[V][T]\to\Bbb{C}[V][y/x]$ is precisely $(xT -y, T^2 - x - 1),$ or by writing $x$ and $y$ in terms of $\Theta$ and justifying that $\Theta$ satisfies no additional relations.
First, note that $\Bbb{C}[V]\cong\Bbb{C}[x,y]/(x^3 + x^2 - y^2)$ and observe that $\Theta^2 = \frac{y^2}{x^2} = \frac{x^3 + x^2}{x^2} = x + 1.$
Now, using the fact that $x = \Theta^2-1,$ we find
\begin{align*}
\Bbb{C}[V][\Theta]&\cong(\Bbb{C}[x,y]/(x^3 + x^2 - y^2))[\Theta]/(x\Theta - y)\\
&= \Bbb{C}[x,y,\Theta]/(x\Theta - y,x^3 + x^2 - y^2)\\
&= \Bbb{C}[y,\Theta]/((\Theta^2 - 1)\Theta - y,(\Theta^2 - 1)^3 + (\Theta^2 - 1)^2 - y^2)
\end{align*}
However, it is now clear that $y = \Theta^3 - \Theta,$ and hence that
\begin{align*}
(\Theta^2 - 1)^3 + (\Theta^2 - 1)^2 - y^2 &=(\Theta^2 - 1)^3 + (\Theta^2 - 1)^2 - (\Theta^3 - \Theta)^2\\
&= (\Theta^2 - 1)^2(\Theta^2 - 1 + 1) - (\Theta^3 - \Theta)^2\\
&= \Theta^2(\Theta^2 - 1)^2 - (\Theta^3 - \Theta)^2\\
&= 0.
\end{align*}
As such, we find that $$((\Theta^2 - 1)\Theta - y,(\Theta^2 - 1)^3 + (\Theta^2 - 1)^2 - y^2) = ((\Theta^2 - 1)\Theta - y),$$ so that
\begin{align*}
\Bbb{C}[V][\Theta]&\cong \Bbb{C}[y,\Theta]/((\Theta^2 - 1)\Theta - y,(\Theta^2 - 1)^3 + (\Theta^2 - 1)^2 - y^2)\\
&=\Bbb{C}[y,\Theta]/((\Theta^2 - 1)\Theta - y)\\
&=\Bbb{C}[\Theta^3 - \Theta,\Theta]\\
&= \Bbb{C}[\Theta].
\end{align*}
A polynomial ring in one variable over a field is clearly a UFD.
A: We have $B=\mathbb C[\bar x,\bar y,\frac{\bar y}{\bar x}]=\mathbb C[\bar x,\frac{\bar y}{\bar x}]=\mathbb C[\frac{\bar y}{\bar x}]$, since $\bar x=(\frac{\bar y}{\bar x})^2-1$.
Now the only thing to prove is that $\frac{\bar y}{\bar x}$ is algebraically independent over $\mathbb C$. If $f\in\mathbb C[t]$, $f=a_0+a_1t+\dots+a_nt^n$ is such that $f(\frac{\bar y}{\bar x})=0$, then $\sum_{i=0}^na_i\bar x^{n-i}\bar y^i=0$. Removing the residue classes we get $\sum_{i=0}^na_ix^{n-i}y^i\in(y^2-x^2-x^3)$. It is easily seen that this holds if and only if $a_i=0$ for all $i$.
